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Serenitypaige1129

  • 2 years ago

What is a quartic function with only the two real zeroes given? x = –1 and x = –3

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  1. tcarroll010
    • 2 years ago
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    You can go with (x + 1)^2 times (x + 3)^2 The reason this works is that it will give you a x^4 term and you still have only the 2 roots. Roots are of the form: (x - root) for you factors.

  2. tcarroll010
    • 2 years ago
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    So, basically you have both roots twice, so that's why each factor has a 2 on it for an exponent.

  3. tcarroll010
    • 2 years ago
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    You could also have put a 3 on one of them and left the other root as a single power (no exponent)

  4. tcarroll010
    • 2 years ago
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    Is this making sense to you?

  5. Serenitypaige1129
    • 2 years ago
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    I am lost completely

  6. tcarroll010
    • 2 years ago
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    np, I'll give you a little more detail: First off, a quartic polynomial function has as its highest exponent, an x^4 term. That means that if you have an equation of the form: y = (x - root1)(x - root2)(x - root3)(x - root4) you will have quartic function. But we can have only 2 different roots, so a good choice is to have root1 twice and root 2 twice: y = [(x - root1)^2][(x - root2)^2]

  7. tcarroll010
    • 2 years ago
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    That's the same as: y = (x - root1)(x - root1)(x - root2)(x - root2) which also has root1 twice and root2 twice. It's just writing out the factors without the exponents. This equation will give you 0 for "y" if you replace x with either root, which is what you want to solve this problem.

  8. tcarroll010
    • 2 years ago
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    And to finish this off, you are given x = -1 and x = -3 as the only 2 roots you can have. And since the factor to give "0" using the root is: (x - root) one factor is [x - (-1)] which is (x + 1) the other is [x - (-3)] which is (x + 3) So, your polynomial is: y = (x + 1)(x + 1)(x + 3)(x + 3)

  9. tcarroll010
    • 2 years ago
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    You have other choices, but the above polynomial works. You could also have: y = (x + 1)(x + 3)(x + 3)(x + 3) or y = (x + 1)(x + 1)(x + 1)(x + 3) Both of those are quartics having only 2 different roots. That should make it all clearer now. Now are you good on this?

  10. tcarroll010
    • 2 years ago
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    Here's a graph of the first solution

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  11. Serenitypaige1129
    • 2 years ago
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    sorta so which one would it be? y = x4 – 4x3 – 4x2 – 4x – 3 y = –x4 + 4x3 + 4x2 + 4x + 3 y = x4 + 4x3 + 3x2 + 4x – 4 y = x4 + 4x3 + 4x2 + 4x +3 I understand how you did that i just don't know which one it will be.

  12. Serenitypaige1129
    • 2 years ago
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    @tcarroll010

  13. tcarroll010
    • 2 years ago
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    I didn't know that you had choices. That changes the whole way you approach this problem. You can solve it 2 ways that I know of. The first way rst and the best and fastest is to perfrom synthetic division on each choice until you hit upon the choice that has only those 2 roots.

  14. tcarroll010
    • 2 years ago
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    Number #1 is out because -1 is not a root. So, I eliminated that one so far.

  15. tcarroll010
    • 2 years ago
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    Number 2 is out because it also has +1 as a root.

  16. tcarroll010
    • 2 years ago
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    Are you sure you wrote those choices down correctly?

  17. Serenitypaige1129
    • 2 years ago
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    I actually copied and paste so yeah

  18. tcarroll010
    • 2 years ago
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    And the roots are supposed to be -1 and -3 ?

  19. Serenitypaige1129
    • 2 years ago
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    yeah that is why i couldn't figure it out. That is what confused me

  20. tcarroll010
    • 2 years ago
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    It looks like like the last one works. I tried it again.

  21. Serenitypaige1129
    • 2 years ago
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    okay thanks again

  22. tcarroll010
    • 2 years ago
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    you're welcome.

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